Friday, May 6, 2016

Proton charge radius

A new user at Physics Stack Exchange, "dandb", has made an observation which I express as follows:

"The charge radius of the proton (in muonic hydrogen) is almost exactly four times the reduced Compton wavelength of the proton."


  1. I made this response at Stack Exchange (possible in correctly as an Answer rather than a comment) and I am reposting it here in case it is deleted:

    Another attractive feature of this conjecture is that it is similar to another conjecture related to hadrons that is known to be true: that the spin of a hadron is equal to the sum of the spins of the quarks in the hadron (which come in discrete half integer increments), even though non-quark partons in the hadron have non-zero spins that "magically" cancel out in the total for reasons that are not well understood (i.e. the "proton spin crisis")). Until we understand why this is the case for spin in hadrons, we can't rule out that this conjecture is exactly true for related reasons.

    I'd also note that the in lots of other areas of the Standard Model (e.g. some of the more obscure relationships between electroweak constants in the Standard Model), there are lots of known exact relationships between Standard Model, so it wouldn't be a priori unreasonable to wonder if there was such a relationship here, particularly given that charge radius is an electroweak phenomenon. I don't think that there is any known Standard Model constant relationship that can explain why this conjecture should be true, but the precision of the the conjectured relationship is sufficiently great that it isn't unreasonable to entertain the possibility that it is exactly or exactly subject to small corrections, true.

    1. I think the thing to do is to calculate charge radius and Compton radius for various charged solitons. If you can find one with this same ratio, you may have hit upon the structure of the proton!

    2. Given the huge variety of different kinds of partons that pop out of protons in its pdf I don't think this is going to get you to the right answer.

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